Problem: Simplify the following expression: $y = \dfrac{7x^2- 38x+15}{7x - 3}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(7)}{(15)} &=& 105 \\ {a} + {b} &=& &=& {-38} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $105$ and add them together. The factors that add up to ${-38}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-3}$ and ${b}$ is ${-35}$ $ \begin{eqnarray} {ab} &=& ({-3})({-35}) &=& 105 \\ {a} + {b} &=& {-3} + {-35} &=& -38 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({7}x^2 {-3}x) + ({-35}x +{15}) $ Factor out the common factors: $ x(7x - 3) - 5(7x - 3)$ Now factor out $(7x - 3)$ $ (7x - 3)(x - 5)$ The original expression can therefore be written: $ \dfrac{(7x - 3)(x - 5)}{7x - 3}$ We are dividing by $7x - 3$ , so $7x - 3 \neq 0$ Therefore, $x \neq \frac{3}{7}$ This leaves us with $x - 5; x \neq \frac{3}{7}$.